where \(\Gamma (z)\)  is Euler's gamma-function [1]}, Vol. I. Moreover, the Parseval equality for the Mellin transform [2]} permits to write (1.5) as the Laplace integral representation for this function. In fact, we obtain
\(\rho _\nu (x)= \int _0^\infty y^{\nu -1} e^{-y - x/y} dy,\quad x >0,\ \nu \in \mathbb {R}.\qquad \mathrm {(1.4)}\) 
